What is the extraneous solution to these equations? $\dfrac{x^2 + x}{x - 3} = \dfrac{12}{x - 3}$
Solution: Multiply both sides by $x - 3$ $ \dfrac{x^2 + x}{x - 3} (x - 3) = \dfrac{12}{x - 3} (x - 3)$ $ x^2 + x = 12$ Subtract $12$ from both sides: $ x^2 + x - (12) = 12 - (12)$ $ x^2 + x - 12 = 0$ Factor the expression: $ (x - 3)(x + 4) = 0$ Therefore $x = 3$ or $x = -4$ At $x = 3$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 3$, it is an extraneous solution.